Question

N one region, the september energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kwh and a standard deviation of 218 kwh. for a randomly selected home, find the probability that the september energy consumption level is between 1100 kwh and 1225 kwh.

Answer 1

The probability that a randomly sample of a normally distributed data with a mean, μ, and standard deviation, σ, is between two numbers (a, b) is given by:

$P(a\ \textless \ X\ \textless \ b)=P(X\ \textless \ b)-P(X\ \textless \ a) \\ \\ =P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)$

Given that the september energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kwh and a standard deviation of 218 kwh.

The probability that the september energy consumption level of a randomly selected home is between 1100 kwh and 1225 kwh is given by:

$P(1100\ \textless \ X\ \textless \ 1225)=P\left(z\ \textless \ \frac{1225-1050}{218} \right)-P\left(z\ \textless \ \frac{1100-1050}{218} \right) \\ \\ =P(z\ \textless \ 0.8028)-P(z\ \textless \ 0.2294)=0.78894-0.5907=\bold{0.1982}$